Question: A particle moves along the curve $xy+y^2=8$ so that that the $x$ -coordinate is increasing at a constant rate of $6$ units per minute. What is the magnitude (in units per minute) of the particle's velocity vector when the particle is at the point $(2,2)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $3\sqrt{29}$ (Choice B) B $2\sqrt{10}$ (Choice C) C $6\sqrt{3}$ (Choice D) D $\dfrac{9}{2}$
Background When working with motion along a curve, we should remember that this motion can also be represented by a position vector $(x,y)$. The main difference is that we're not given the equations for $x$ and $y$ in terms of time $t$, but only the relationship between $x$ and $y$ themselves. This however shouldn't prevent us from assuming the expressions for $x$ and $y$ in terms of $t$ exist. As always, $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$. Setting up the math We are given that $\dfrac{dx}{dt}=6$ for any value of $t$. We are asked for the magnitude of the particle's velocity vector when the particle is at the point $(2,2)$. In other words, we are asked for $\left|\left|\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)\right|\right|$ at the point $(2,2)$. Finding $\dfrac{dy}{dt}$ $\dfrac{dy}{dt}=-\dfrac{6y}{x+2y}$ Finding $\dfrac{dy}{dt}$ at $(2,2)$ The expression for $\dfrac{dy}{dt}$ depends on both the particle's $x$ -coordinate ${2}$ and its $y$ -coordinate ${2}$ : $\begin{aligned} \dfrac{dy}{dt}&=-\dfrac{6({2})}{({2})+2({2})} \\\\ &=-\dfrac{12}{6} \\\\ &=-2 \end{aligned}$ Therefore, the particle's velocity vector at the point $(2,2)$ is $(6,-2)$. Finding $||(6,-2)||$ $||(6,-2)||=2\sqrt{10}$ In conclusion, the magnitude of the particle's velocity vector when the particle is at the point $(2,2)$ is $2\sqrt{10}$ units per minute.